I like this as it avoids $\delta, \epsilon$. :)
–
gnometoruleJan 8 '13 at 2:33

@gnometorule $\delta, \epsilon$ was my first instinct actually, but then it got ugly arguing about the middle like the above solutions, so I decided to try modifying the splitting trick.
–
Calvin LinJan 8 '13 at 2:36

This is a very neat argument. +1. You could directly conclude that $$\sum \dfrac{(a_i-a)(b_i-b)}n \to 0$$ since $(a_i-a)(b_i-b) \to 0$, just like you have concluded $\sum \dfrac{a_i}n \to a$ and $\sum \dfrac{b_i}n \to b$.
–
user17762Jan 8 '13 at 2:40

@Marvis Due to the indices being $i$ and $n-i$, I'm not certain / immediately obvious that we can switch the summation signs. We know that $\lim_{n \rightarrow \infty} (a_i - a)(b_{n-i} - b) = 0$, but can't do the sum the other way.
–
Calvin LinJan 8 '13 at 2:49

The idea is to consider only the middle. Using the definition of convergence, pick some large $m$, such that $a_k, k > m$ is almost $a$ and $b_k, k > m$ is alomst $b$. Take $n$ much larger than $m$. Then you have (1) $n-2m$ terms that are almost $ab$, (2) some constant to represent the remaining $2m$ terms and (3) something small to represent the almost. (1) goes to $ab$ as $n \to \infty$. (2) goes to $0$ as well. (3) goes to $0$ as $m$ goes to infinity. So essentially you have two limits now.

Hint: subtract $-ab$ from the left side, gathering terms as $t_i := a_i b_{n-i} - ab$ on numerator, and show it goes to $0$. To do this, note (i) there is a $k_0$ such that for $i, n-i > k_0$ the term $t_i$ will be smaller than an $\epsilon$ you choose, (ii) there will be $n - 2k$ such terms smaller than $\epsilon$, and $\frac {n - 2k}{n}$ goes to 1, and (iii) the $k_0$ terms each "on the left and right" of what we control in (i) and (ii) go to $0$ as they can be appropriately bounded in the numerator, and are divided by $n$.

This is a generalization of a Cesaro mean which can probably be found somewhere on MSE.
Admitting this, we know that
$$
\lim_{n\rightarrow +\infty}\frac{b_1+\cdots+b_n}{n}=\lim_{n\rightarrow +\infty}\frac{B_n}{n}=b.
$$

Now it only remains to show that $\lim_{n\rightarrow +\infty} v_n=0$ where
$$
v_n=u_n-a\cdot\frac{B_n}{n}=\frac{a_1b_n+\cdots+a_nb_1}{n}-\frac{ab_n+\cdots+ab_1}{n}=\frac{1}{n}\sum_{k=1}^{n}(a_k-a)b_{n-k+1}.
$$

Since $b_n$ converges, $|b_n|$ is bounded, say, by $M$.
Then we have
$$
|v_n|\leq \frac{M}{n}\sum_{k=1}^{n}|a_k-a|
$$
for all $n$.
Another application of the Cesaro mean shows that the RHS converges to $0$, and so does $v_n$.